3.13.6 \(\int \frac {x^{11}}{\sqrt [4]{a-b x^4}} \, dx\) [1206]

Optimal. Leaf size=62 \[ -\frac {a^2 \left (a-b x^4\right )^{3/4}}{3 b^3}+\frac {2 a \left (a-b x^4\right )^{7/4}}{7 b^3}-\frac {\left (a-b x^4\right )^{11/4}}{11 b^3} \]

[Out]

-1/3*a^2*(-b*x^4+a)^(3/4)/b^3+2/7*a*(-b*x^4+a)^(7/4)/b^3-1/11*(-b*x^4+a)^(11/4)/b^3

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Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {272, 45} \begin {gather*} -\frac {a^2 \left (a-b x^4\right )^{3/4}}{3 b^3}-\frac {\left (a-b x^4\right )^{11/4}}{11 b^3}+\frac {2 a \left (a-b x^4\right )^{7/4}}{7 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^11/(a - b*x^4)^(1/4),x]

[Out]

-1/3*(a^2*(a - b*x^4)^(3/4))/b^3 + (2*a*(a - b*x^4)^(7/4))/(7*b^3) - (a - b*x^4)^(11/4)/(11*b^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^{11}}{\sqrt [4]{a-b x^4}} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{a-b x}} \, dx,x,x^4\right )\\ &=\frac {1}{4} \text {Subst}\left (\int \left (\frac {a^2}{b^2 \sqrt [4]{a-b x}}-\frac {2 a (a-b x)^{3/4}}{b^2}+\frac {(a-b x)^{7/4}}{b^2}\right ) \, dx,x,x^4\right )\\ &=-\frac {a^2 \left (a-b x^4\right )^{3/4}}{3 b^3}+\frac {2 a \left (a-b x^4\right )^{7/4}}{7 b^3}-\frac {\left (a-b x^4\right )^{11/4}}{11 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 40, normalized size = 0.65 \begin {gather*} \frac {\left (a-b x^4\right )^{3/4} \left (-32 a^2-24 a b x^4-21 b^2 x^8\right )}{231 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a - b*x^4)^(1/4),x]

[Out]

((a - b*x^4)^(3/4)*(-32*a^2 - 24*a*b*x^4 - 21*b^2*x^8))/(231*b^3)

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Maple [A]
time = 0.15, size = 37, normalized size = 0.60

method	result	size
gosper	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 b^{2} x^{8}+24 a b \,x^{4}+32 a^{2}\right )}{231 b^{3}}\)	\(37\)
trager	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 b^{2} x^{8}+24 a b \,x^{4}+32 a^{2}\right )}{231 b^{3}}\)	\(37\)
risch	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 b^{2} x^{8}+24 a b \,x^{4}+32 a^{2}\right )}{231 b^{3}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(-b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-1/231*(-b*x^4+a)^(3/4)*(21*b^2*x^8+24*a*b*x^4+32*a^2)/b^3

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Maxima [A]
time = 0.29, size = 50, normalized size = 0.81 \begin {gather*} -\frac {{\left (-b x^{4} + a\right )}^{\frac {11}{4}}}{11 \, b^{3}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} a}{7 \, b^{3}} - \frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

-1/11*(-b*x^4 + a)^(11/4)/b^3 + 2/7*(-b*x^4 + a)^(7/4)*a/b^3 - 1/3*(-b*x^4 + a)^(3/4)*a^2/b^3

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Fricas [A]
time = 0.36, size = 36, normalized size = 0.58 \begin {gather*} -\frac {{\left (21 \, b^{2} x^{8} + 24 \, a b x^{4} + 32 \, a^{2}\right )} {\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{231 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

-1/231*(21*b^2*x^8 + 24*a*b*x^4 + 32*a^2)*(-b*x^4 + a)^(3/4)/b^3

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Sympy [A]
time = 0.47, size = 70, normalized size = 1.13 \begin {gather*} \begin {cases} - \frac {32 a^{2} \left (a - b x^{4}\right )^{\frac {3}{4}}}{231 b^{3}} - \frac {8 a x^{4} \left (a - b x^{4}\right )^{\frac {3}{4}}}{77 b^{2}} - \frac {x^{8} \left (a - b x^{4}\right )^{\frac {3}{4}}}{11 b} & \text {for}\: b \neq 0 \\\frac {x^{12}}{12 \sqrt [4]{a}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(-b*x**4+a)**(1/4),x)

[Out]

Piecewise((-32*a**2*(a - b*x**4)**(3/4)/(231*b**3) - 8*a*x**4*(a - b*x**4)**(3/4)/(77*b**2) - x**8*(a - b*x**4
)**(3/4)/(11*b), Ne(b, 0)), (x**12/(12*a**(1/4)), True))

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Giac [A]
time = 0.63, size = 57, normalized size = 0.92 \begin {gather*} -\frac {21 \, {\left (b x^{4} - a\right )}^{2} {\left (-b x^{4} + a\right )}^{\frac {3}{4}} - 66 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{231 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

-1/231*(21*(b*x^4 - a)^2*(-b*x^4 + a)^(3/4) - 66*(-b*x^4 + a)^(7/4)*a + 77*(-b*x^4 + a)^(3/4)*a^2)/b^3

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Mupad [B]
time = 1.16, size = 38, normalized size = 0.61 \begin {gather*} -{\left (a-b\,x^4\right )}^{3/4}\,\left (\frac {32\,a^2}{231\,b^3}+\frac {x^8}{11\,b}+\frac {8\,a\,x^4}{77\,b^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(a - b*x^4)^(1/4),x)

[Out]

-(a - b*x^4)^(3/4)*((32*a^2)/(231*b^3) + x^8/(11*b) + (8*a*x^4)/(77*b^2))

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